Product/quotient of factorials beget dyadic powers

Question

I am writing up some notes and the following occurred to me and I would like to see if there are a variety of ways to prove it. Just for reference, the identity pops out of equality between constant term evaluations of Laurent polynomials from my earlier quest. There should be easier ways as well, I think.

QUESTION. Can you provide a proof for $$\prod_{j=0}^n\frac{(2j)!^2}{(n+j)!\,j!}=2^n.$$

Answer 1

This is easy. Start with $$n!.2^n=(2n)(2n-2)\dots 2.$$ This implies that $$((2n-1)!(2n-3)!\dots 1!).n!.2^n=(2n)!(2n-2)!\dots 2!.$$ Then $$((2n)!(2n-1)!\dots 1!).n!.2^n=((2n)!(2n-2)!\dots 2!)^2.$$ Thus $$\left(\prod_{j=0}^n (n+j)!j!\right).2^n=\prod_{j=0}^n (2j)!^2. $$ Finally, divide both sides by the product on the left.